I'm trying to find the maximum bending moment and deflection for this symmetrical simple beam. The formulae I can find, deal with either cantilever beams with one fixed end, or continuous beams without cantilever ends, not this kind of situation.

enter image description here

I've tried to model the beam as two fixed-end cantilevers, on the basis that symmetry demands the middle of the beam is horizontal, and then modelled the loads as separate loads superimposed. But I'm not getting anywhere and not sure how I'd tell whether my answer was correct, if I did. I don't know if the fact that the beam can rise or fall in the middle (depending how F compares to W) is the problem, or if it's something else.

The beam won't be exposed to non-linear behaviour. It could bend either way, as I don't know whether the 2 point forces will be larger or smaller than the UDL, or the exact lengths involved.

Is there an easy way to work out the maximum deflection and B.M. of this beam, and if so, what are they?

Note, this isn't homework or coursework (I haven't had those for many years!). It's part of a real-world practical design I'm trying to work on.


3 Answers 3


You are correct that symmetry demands that rotation at midspan be zero. However, as you also mentioned, there will obviously be deflection at midspan as well. You therefore cannot simplify the structure by placing a fixed support at midspan (not entirely sure what you meant by "two fixed-end cantilevers", since you'd obviously still need to consider the pinned support).

Instead, you need to simplify the structure by putting a "fixed rotation" support. That is, one which only resists rotation, presenting no force reactions, just a bending moment reaction.

That being said, you seem to be trying to do this from tables, without going into the math. In that spirit, let's see what we can do. For the record, I don't have a table handy, so I did some googling and will be using this one.

For starters, let's make use of the superposition principle and look at the loads individually (or rather, look at the distributed load first and then the pair of concentrated forces later).

Let's start with the distributed load:

Calculating the results for the midspan is straightforward. Just pretend this is a simply supported beam with no cantilevers (obviously this only works because the distributed load is only on the span and not on the cantilevers).

Looking at tables, we know that the bending moment and deflection at midspan are

$$\begin{align} M_{midspan} &= \dfrac{wL^2}{8} \\ \delta_{midspan} &= -\dfrac{5wL^4}{384EI} \end{align}$$

(Note I'm using $w$ and later $F$ positive in the downward direction, feel free to invert the signs if you prefer another convention)

But what about the cantilevers? Well, we trivially know that $M_{support} = 0$ since the entirety of the distributed load is absorbed by the supports, so there's nothing on the cantilevers to generate any moment.

However, the cantilevers will present a deflection due to the rotation at the support. Though the lack of loading means that the cantilevers will act as undeformed straight beams, they will suffer a rigid-body rotation which will lead to a deflection at the end.

Basic trigonometry tells us that the vertical deflection of the cantilever will be equal to

$$\delta_{cant} = d\cdot\tan\theta_{support}$$

However, civil engineers are lazy, so we just take the first-order Taylor series of $\tan\theta$, and simplify it to merely $\theta$ (this is only valid for small deflections and rotations, but that's the usual space inhabited by civil engineering, so it's fine for our purposes). So we can simplify the above to merely

$$\delta_{cant} = d\theta_{support}$$

In this case, $\theta_{support}$ can also be found in tables, and is equal to

$$\begin{align} \theta_{support} = \dfrac{wL^3}{24EI} \\ \therefore \delta_{cant} = d\dfrac{wL^3}{24EI} \end{align}$$

Now let's move onto the concentrated forces:

We know that the bending moment diagram for these forces will look like this:

enter image description here

We can trivially observe that $M_{support} = -Fd$ (negative because it generates tension on the top fibers). The constant bending moment diagram along the span tells us that the totality of each concentrated load is entirely absorbed by the nearest support. This means that, as far as the span is concerned, there are no concentrated forces at all! It's only being acted upon by a constant bending moment.

Therefore, we can change the model (when looking at the midspan) into this:

enter image description here

Thankfully, this case is present in many tabled results. We already know the bending moment, but we're still looking for the deflection.

My table only has the result for a single concentrated bending moment at one end, so we'll just have to double the deflection it gives us for the midspan (which we know is where the maximum deflection actually happens in our case):

$$\delta_{midspan} = 2\times\dfrac{ML^2}{16EI} = \dfrac{FdL^2}{8EI}$$

So with that we have the bending moment diagram due to the concentrated loads and the maximum deflection at midspan due to those loads at the cantilever.

What we don't yet have, however, is the deflection at the cantilevers themselves.

For that, we'll need to get a bit creative.

The deflection at the cantilevers is a function of two things: the rotation at the support and the additional deformation which happens between the support and the cantilever's end. Thankfully, both of those things are easy to find.

The rotation at the support can be found using the same concentrated-bending-moments model we used to find the deflection at midspan, and then looking at the same table. In this case, the table gives us the rotation at the support with the bending moment and at the other. It also gives them different signs since the beam has a positive slope on one end and a negative slope on the other. In our case, we're just going to add both of those values ignoring the signs, since our double-bending-moment case will increase the rotation at both supports.

$$\theta_{support} = \dfrac{ML}{6EI} + \dfrac{ML}{3EI} = \dfrac{ML}{2EI}$$

Now, just as we did with the distributed load, the deflection at the cantilever due to this rotation at the support can be easily found:

$$\delta_{cant,rotation} = d\theta_{support}$$

As for the additional deflection due to the cantilever's deformation after the support, we can simply model the cantilever alone as if it had a fixed support instead of a pinned one.

The table I used before doesn't actually have cantilever results, but I remember that a cantilever's maximum deflection is equal to

$$\delta_{cant,deform} = \dfrac{Fd^3}{3EI}$$

We can therefore conclude that the total deformation at the cantilever due to the concentrated loads is equal to

$$\begin{align} \delta_{cant} &= d\theta_{support} + \dfrac{Fd^3}{3EI} \\ &= d\dfrac{ML}{2EI} + \dfrac{Fd^3}{3EI} \\ &= d\dfrac{FdL}{2EI} + \dfrac{Fd^3}{3EI} \\ &= \dfrac{F}{EI}\left(\dfrac{d^2L}{2} + \dfrac{d^3}{3}\right) \\ \end{align}$$

After all that, we now bring everything together by adding the results from the distributed load and concentrated forces:

$$\begin{align} M_{midspan} &= \dfrac{wL^2}{8} - Fd \\ \delta_{midspan} &= -\dfrac{5wL^4}{384EI} + \dfrac{FdL^2}{8EI} \\ M_{support} &= 0 - Fd \\ \delta_{cant} &= \dfrac{wdL^3}{24EI} - \dfrac{F}{EI}\left(\dfrac{d^2L}{2} + \dfrac{d^3}{3}\right) \end{align}$$

  • $\begingroup$ Wow. Just.. Wow. Thank you for an answer that I can not just use, but work from if I need other solutions like it! $\endgroup$
    – Stilez
    Mar 27, 2019 at 18:43

I don't think we can have an exact solution or even a proper estimate without having some numbers. If you look at the BM diagram of the structure, it looks something like this: enter image description here

Now, it is quite obvious that without knowing the lengths and the load values involved we will never be able to estimate which peak will be higher. Similarly, for deflection, the magnitudes of the load affect the deflection pattern.

  • $\begingroup$ This is actually pretty helpful - thank you! $\endgroup$
    – Stilez
    Mar 27, 2019 at 18:43


After drawing Free Body, Bending moment $ M/EI $ diagrams, integrations by

First & Second Moment Area Methods

help determine rotations and deflections.


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